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I can't seem to find the proper integrating factor for this nonlinear first order ODE. I have even tried pulling a bunch of substitution and equation-manipulating tricks, but I can't seem to get a proper integrating factor.

$$frac{1}{x}dx + left(1+x^2y^2right)dy = 0$$

EDIT: Due to MSE users complaining about my lack of proof of work, intent of conceptual understanding, etc, here is exactly why I am stuck.

To start off, this ODE is obviously inexact:

$$frac{partial}{partial y}left(frac{1}{x}right) neq frac{partial}{partial x}left(1+x^2y^2right)$$

And so in order to make this exact (if we choose to go down this route) we must (I'll stick to standard convention/notation) find a function $mu$ such that if we multiply the entire original ODE by it, we will be able to integrate and solve using 'exact ODE' methods. This is shown as:

$$ mu left(frac{1}{x}right)dx + mu left(1+x^2y^2right)dy = 0$$

$$ frac{partial}{partial y} left(muleft(frac{1}{x}right) right) = frac{partial}{partial x} left(mu left(1+x^2y^2right) right)$$

Now expanding by chain rule, we get:

$$mu_y left(frac{1}{x}right) = mu_x left(1+x^2y^2right) + mu left(2xy^2right)$$

Now here is where I'm stuck. We want to avoid dealing with a PDE, so we try to stick to good old ODE techniques by assuming that $mu$ is either a function of only x or only y.
Let's first assume that $mu$ is only a function of y. The following will then be true.

$$ mu_x = 0$$

$$ mu_y left(frac{1}{x} right) = mu left(2xy^2 right)$$

$$ frac{dmu}{mu} = 2x^2y^2 dy$$

By looking at the right hand side, we see that it just won't work - x and y are related, so we can't have that integral.

Now let's assume that $mu$ is only a function of x. The following will then be true.

$$ mu_y = 0$$

$$ mu_x left(1+x^2y^2right) = -mu left(2xy^2right)$$

$$ frac{dmu}{mu} = frac{-2xy^2}{1+x^2y^2} dx$$

And, once again, if you look at the right hand side, we have an integral that we can't immediately work out, just as in the previous case.

It looks to me like we can't find an integrating factor which depends only on $x$ or only on $y$ (in general, $mu$ will be a function of only one variable just in some special cases, so this is not entirely surprising).

For any equation of the form $p(x,y)dx + q(x,y)dy = 0$,
in order to be able to find $mu := mu(x)$ it must be the case that
begin{equation}
frac{frac{partial p}{partial y} - frac{partial q}{partial x}}{q}
end{equation}
is a function of $x$ only. If it is, we can set
begin{equation}
mu(x) = expleft(intfrac{frac{partial p}{partial y} - frac{partial q}{partial x}}{q}dxright).
end{equation}
Here I get
begin{equation}
frac{frac{partial p}{partial y} - frac{partial q}{partial x}}{q} =
frac{-2xy^2}{1 + x^2y^2},
end{equation}
which clearly depends on $y$. Then we can't have a $mu$ which depends only on $x$. To have a $mu$ which depends only on $y$, we must have
begin{equation}
frac{frac{partial q}{partial x} - frac{partial p}{partial y}}{p}
end{equation}
be only a function of $y$. If that's the case, we can set
begin{equation}
mu(y) = expleft(int frac{frac{partial q}{partial x} - frac{partial p}{partial y}}{p} dyright).
end{equation}
Here I get
begin{equation}
frac{frac{partial q}{partial x} - frac{partial p}{partial y}}{p} = frac{2xy^2}{frac{1}{x}} = 2x^2y^2,
end{equation}
which clearly depends on $x$. Then we can't have $mu$ depend on $y$ only.

As a result, we must have $mu$ depend on both $x$ and $y$.

Assume you're looking for a solution $y(x)$, then mathematic gives:

$$ frac{-1}{x + x^3 y^2} = frac{ dy}{dx} implies frac{exp (-2 y)}{8 x^2} + frac{ exp (-2y) }{16} (2 y^2 +2y +1) = Const $$